Question

The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths and img src="https://tex.z-dn.net/?f=%242%5Csqrt%7B3%7D%2C~5%2C%24%20and%20%24%5Csqrt%7B37%7D%2C%24%20" id="TexFormula1" title="$2\sqrt{3},~5,$ and $\sqrt{37},$ " alt="$2\sqrt{3},~5,$ and $\sqrt{37},$ " align="absmiddle" class="latex-formula">as shown, is $\frac{m\sqrt{p}}{n},$ where m, n, and p are positive integers, m and n are relatively prime, and p is not divisible by the square of any prime. What is m + n + p?

Answer 1

Set up the given triangle on x-y coordinates with right angle at (0,0). So the two vertices are at (5,0) and (0,2$sqrt{x} n]{3}$)

let (a,0) and (0,b) be two vertices of the equilateral triangle. So the third vertex must be at $(\frac{a+\sqrt{3}b}{2} , \frac{b+\sqrt{3}a}{2} )$

for a pt (x,y) on line sx+ty=1, the minimum of $\sqrt{x^{2} + {y^{2} }$
equals to $\frac{1}{ \sqrt{ s^{2} + t^{2} } }$

smallest value happens at $\frac{10 \sqrt{3} }{67}$

so area is $\frac{75\sqrt{3}}{67}$

hence m=75, n=67, p=3
m+n+p = 75+67+3 = 145

Answer 2

Hey there,

Your correct answer final answer to this question would be 145.

Hope this helps.

~Jurgen